Peak values for a neodymium-glass laser are 600 J for 1.00 nanosecond.If the cross-section of the laser beam is 1.10 cm²,what are the maximum values of E and B? (µ₀ = 4π × 10−⁷ T⋅m/A,c = 3.00 × 10⁸ m/s) A)2.0E+9 V/m,1.9 T B)5.7E+8 V/m,6.8 T C)2.0E+9 V/m,6.8 T D)5.7E+8 V/m,1.9 T E)Not enough information.

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Quizplus · Accepted Answer

The peak electric field is:$$E=\frac{P}{Ac}=\frac{600	ext{ J}}{(1.10	imes10^{-4}	ext{ m}^2)(3.00	imes10^8	ext{ m/s})}=2.0	imes10^9	ext{ V/m}$$The peak magnetic field is:$$B=\frac{E}{c}=\frac{2.0	imes10^9	ext{ V/m}}{3.00	imes10^8	ext{ m/s}}=6.8	ext{ T}$$

Peak Values for a Neodymium-Glass Laser Are 600 J for 1.00